Page 150 - MarceAlgebra Demystified

CHAPTER 6 Factoring 137

2
2: x þ 2x þ 1 ¼
2
3: x þ 3x 10 ¼
2
4: x 6x þ 8 ¼
2
5: x 11x 12 ¼
2
6: x 9x þ 14 ¼

2
7: x þ 7x þ 10 ¼
2
8: x þ 4x 21 ¼

Solutions

2
1: x 5x 6 ¼ðx Þðx þ Þ
2
2: x þ 2x þ 1 ¼ðx þ Þðx þ Þ

2
3: x þ 3x 10 ¼ðx Þðx þ Þ
2
4: x 6x þ 8 ¼ðx Þðx Þ
2
5: x 11x 12 ¼ðx Þðx þ Þ
2
6: x 9x þ 14 ¼ðx Þðx Þ
2
7: x þ 7x þ 10 ¼ðx þ Þðx þ Þ
2
8: x þ 4x 21 ¼ðx Þðx þ Þ
Once the signs are determined all that remains is to fill in the two blanks.
Look at all of the pairs of factors of the constant term. These pairs will be
the candidates for the blanks. For example, if the constant term is 12, you
will need to consider 1 and 12, 2 and 6, and 3 and 4. If both signs in the
factors are the same, these will be the only ones you need to try. If the
signs are different, you will need to reverse the order: 1 and 12 as well as
12 and 1; 2 and 6 as well as 6 and 2; 3 and 4 as well as 4 and 3. Try the
FOIL method on these pairs. (Not every trinomial can be factored in this
way.)